b. The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output).

c. Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors.

c. n1 + n2 = 50 large eggs min

b n1 large eggs broken/50 large eggs = 11 50 = 0.22 g d. b g 22% of the large eggs (right hand) and 25 70 ⇒ 36% of the extra-large eggs (left hand) are broken. Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed.

lnbag = lnb x g − b lnb R g = lnb0100

1 1 . lnb15g = −6.340 ⇒ a = 1764 . g − 1491 . × 10 −3

x = 1764 . × 10−3 R1.491

F x I F 0.900 IJ 1 1

R=G J =G b 1.491

H a K H 1764 . × 10 K −3 = 655 .

d. Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass fractions – measure against known standard. Impurities in the stream – analyze a sample. Mixture is not all liquid – check sample. Calibration data are temperature dependent – check calibration at various temperatures. System is not at steady state – take more measurements. Scatter in data – take more measurements.

4.41 a. By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed. Automation provides for faster and more accurate response to fluctuations in the feed stream, reducing the risk of release of H2S and SO2. It also may reduce labor costs.

b. As before, n1 = 160.7 mol air fed , n2 = 15 mol HCl

N 2 : n4 = 0.79b160.7g = 127 mol N 2

Cl 2 : n5 = ξ = 42.5 mol Cl 2 H 2 O: n6 = ξ = 42.5 mol H 2 O

These molar quantities are the same as in part (a), so the mole fractions would also be the same. c. Use of pure O2 would eliminate the need for an extra process to remove the N2 from the product gas, but O2 costs much more than air. The cheaper process will be the process of choice.

d. The regulation was avoided by diluting the stack gas with fresh air before it exited from the stack. The new regulation prevents this since the mass of SO2 emitted per mass of coal burned is independent of the flow rate of air in the stack.

b. If more air is fed to the furnace,

(i) more gas must be compressed (pumped), leading to a higher cost (possibly a larger pump, and greater utility costs) (ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the product gas temperature decreases and less steam is produced.

Condensation measurement: b1.134 g H Ogb1 mol 18.02 gg = 0126 2 . mole H 2 O 0.50 mol product gas mole product gas Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates if it were necessary (which it is not).